# How do you solve root4(2x)+root4(x+3)=0?

Apr 29, 2017

This equation has no solutions.

#### Explanation:

Note that $\sqrt[4]{t} \ge 0$ for all $t \ge 0$ and has no real value when $t < 0$.

So the only circumstances under which the left hand side of the given equation can be $0$ are if both:

$2 x = 0 \text{ }$ and $\text{ } x + 3 = 0$

The first of these implies that $x = 0$ and the second that $x = - 3$.

These are incompatible, so there are no real solutions.

$\textcolor{w h i t e}{}$
Complex solutions?

Given:

$\sqrt[4]{2 x} + \sqrt[4]{x + 3} = 0$

Subtract $\sqrt[4]{x + 3}$ from both sides to get:

$\sqrt[4]{2 x} = - \sqrt[4]{x + 3}$

Raise both sides to the $4$th power (noting that this is where we might introduce extraneous solutions) to get:

$2 x = x + 3$

Subtract $3$ from both sides to find:

$x = 3$

But...

$\sqrt[4]{2 \left(\textcolor{b l u e}{3}\right)} + \sqrt[4]{\textcolor{b l u e}{3} + 3} = \sqrt[4]{6} + \sqrt[4]{6} \ne 0$

So there are no solutions at all.